Below I've posted the code to a non-working "divide and conquer" multiplication method in ruby(with debug prints). I cannot tell if its broken code, or a quirk in Ruby like how the L-shift(<<) operator doesn't push bits into the bit-bucket; this is unexpected compared to similar operations in C++.
Is it broken code (doesn't match the original algorithm) or unexpected behavior?
Pseudo code for original algorithm
def multiply(x,y,n, level)
#print "Level #{level}\n"
if n == 1
#print "\tx[#{x.to_s(2)}][#{y.to_s(2)}]\n\n"
return x*y
end
mask = 2**n - 2**(n/2)
xl = x >> (n / 2)
xr = x & ~mask
yl = y >> (n / 2)
yr = y & ~mask
print " #{n} | x = #{x.to_s(2)} = L[#{xl.to_s(2)}][#{xr.to_s(2)}]R \n"
print " #{n} | y = #{y.to_s(2)} = L[#{yl.to_s(2)}][#{yr.to_s(2)}]R \n"
#print "\t[#{xl.to_s(2)}][#{yr.to_s(2)}]\n"
#print "\t[#{xr.to_s(2)}][#{yr.to_s(2)}]\n"
#print "\t([#{xl.to_s(2)}]+[#{xr.to_s(2)}])([#{yl.to_s(2)}]+[#{yr.to_s(2)}])\n\n"
p1 = multiply( xl, yl, n/2, level+1)
p2 = multiply( xr, yr, n/2, level+1)
p3 = multiply( xl+xr, yl+yr, n/2, level+1)
return p1 * 2**n + (p3 - p1 - p2) * 2**(n/2) + p2
end
x = 21
y = 22
print "x = #{x} = #{x.to_s(2)}\n"
print "y = #{y} = #{y.to_s(2)}\n"
print "\nDC_multiply\t#{x}*#{y} = #{multiply(x,y,8, 1)} \nregular\t#{x}*#{y} = #{x*y}\n\n "
I am not familiar with the divide and conquer algorithm but i don't think it contains parts you can't do in Ruby.
Here is a quick attempt:
def multiplb(a,b)
#Break recursion when a or b has one digit
if a < 10 || b < 10
a * b
else
#Max number of digits of a and b
n = [a.to_s.length, b.to_s.length].max
# Steps to split numbers to high and low digits sub-numbers
# (1) to_s.split('') => Converting digits to string then arrays to ease splitting numbers digits
# (2) each_slice => Splitting both numbers to high(left) and low(right) digits groups
# (3) to_a , map, join, to_i => Simply returning digits to numbers
al, ar = a.to_s.split('').each_slice(n/2).to_a.map(&:join).map(&:to_i)
bl, br = b.to_s.split('').each_slice(n/2).to_a.map(&:join).map(&:to_i)
#Recursion
p1 = multiplb(al, bl)
p2 = multiplb(al + ar, bl + br)
p3 = multiplb(ar, br)
p1 * (10**n) + (p2 - p1 - p3) * (10**(n/2)) + p3
end
end
#Test
puts multiplb(1980, 2315)
# => 4583700 yeah that's correct :)
Here are some references to further explain part of the code:
Finding max of numbers => How do you find a min / max with Ruby?
Spliting an array to half => Splitting an array into equal parts in ruby
Turning a fixnum into array => Turning long fixed number to array Ruby
Hope it hepls !